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Q.
The amount of oxalic acid, required to prepare $300 \,mL \,2.5\, M$ solution, is:
Jharkhand CECEJharkhand CECE 2004
Solution:
Molarity $=\frac{\text { moles of solute }}{\text { volume of solution in L }}=\frac{\text { mass/molecular mass }}{\text { V in L }}$
Given molarity $=2.5\, M$
Volume $=300 \,mL =\frac{300}{1000} L$
Molecular mass of oxalic acid $( COOH )_{2}$
$=(12+16 \times 2+1)_{2}=902.5=\frac{\text { mass } / 90}{300 / 1000}$
or $2.5=\frac{\text { mass }}{90} \times \frac{1000}{300}$
mass $=\frac{90 \times 300 \times 2.5}{1000}=67.5 \,g$